Найдите область определения функции y=корень из (12-8x+x^2).

Ответ:

\[y = \sqrt{12 - 8x + x^{2}}\]

\[12 - 8x + x^{2} \geq 0\]

\[x^{2} - 8x + 12 \geq 0\]

\[D_{1} = 16 - 12 = 4\]

\[x_{1} = 4 - 2 = 2;\ \ x_{2} = 4 + 2 = 6.\]

\[(x - 2)(x - 6) \geq 0\]

\[x \leq 2;\ \ x \geq 6.\]

\[D(y) = ( - \infty;2\rbrack \cup \lbrack 6; + \infty).\]